Question 5

Solution by Jayanta Manoharmayum

Let \(\{p_1,\dots,p_n,...\}\) be the set of all integers of the form \(m^k\) with \(p_i< p_{i+1}\) and \( k>1\). Suppose that there exists an integer \(N\) such that \(p_{i+1}-p_i < N\) for every \(i\). Then \[p_2 < 1+N < 2N,\qquad p_3 < 1+2N < 3N,\qquad \dots\qquad p_n < 1+(n-1)N < nN,\qquad \dots.\] So, \[ \sum_{m=2}^{\infty}\tfrac1{p_i}>\sum_{m=2}^{\infty}\tfrac{1}{mN}= \tfrac1N\sum_{m=2}^{\infty}\tfrac{1}{m} \] which diverges. However, \begin{gather*} \sum_{m=2}^{\infty}\tfrac1{p_i} \leq \left(\tfrac1{2^2}+\tfrac1{2^3}+...\right) + \left(\tfrac1{3^2}+\tfrac1{3^3}+...\right)+...+ \left(\tfrac1{m^2}+\tfrac1{m^3}+...\right)+...\\ = \tfrac1{2^2}\left(1+\tfrac12+...\right)+\tfrac1{3^2}\left(1+\tfrac13+...\right) + ...+ \tfrac1{m^2}\left(1+\tfrac1{m}+...\right)+...\\ = \tfrac1{2^2}\left(\tfrac21\right)+\tfrac1{3^2}\left(\tfrac32\right) + ...+ \tfrac1{m^2}\left(\tfrac{m}{m-1}\right)+...\\ = \sum_{m=2}^{\infty}\tfrac1m\left(\tfrac1{m-1}\right) = \sum_{m=2}^{\infty}\left(\tfrac1{m-1}-\tfrac1m \right) = 1 \end{gather*} a contradiction. Therefore no such \(N\) exists and we conclude that there are arbitrarily large gaps between powers of integers.